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The Binomial & Geometric Distribution

The Binomial & Geometric Distribution. Chapter 8. Everyone ’ s worst nightmare. Answer: C E D E A B C E D A. Activity: In a multiple choice test, there will be five choices for each question: A B C D E. Select an answer to each question. Binomial Coefficient.

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The Binomial & Geometric Distribution

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  1. The Binomial & Geometric Distribution • Chapter 8

  2. Everyone’s worst nightmare • Answer: • C • E • D • E • A • B • C • E • D • A Activity: In a multiple choice test, there will be five choices for each question: A B C D E. Select an answer to each question.

  3. Binomial Coefficient

  4. NOTATION X: The number of successes that result from the binomial experiment.n:The number of trials in the binomial experiment. P: The probability of success on an individual trial.Q: The probability of failure on an individual trial. (This is equal to 1 - P.)b(x; n, P): Binomial probability - nCr: The number of combinations of n things, taken r at a time.

  5. x=2 n=5 p=.167 Binomial Formula b = (x;n,p) = Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours? number of trials is equal to 5 number of successes is equal to 2 the probability of success (getting a 4) on a single trial is 1/6 or about 0.167

  6. x=2 n=5 p=.167 =5C2 (.167)2 * (1-.167)5-2 =10 (.167) 2(.833) 3 =0.161 The Probability of getting exactly two 4’s after 5 trials is 16%

  7. Suppose a coin is tossed 5 times. What is the probability of getting exactly 3 heads? n=5 x=3 p=.5 b=(n,x,p) =5C3 (.5)3 * (1-.5)5-3 =10 (.5)3 * (1-.5)5-3 =.3125

  8. What is the probability of getting 8 correct answers from guessing on the pop quiz? The probability of getting 8 correct answers from guessing on the quiz will be .0000737%

  9. Peter is a basketball player who makes 75% of his free throws over the course of the season. Peter shoots 12 free throws and makes only 7 of them. The fans think he failed because he is nervous. Is it unusual for Peter to perform this poorly? b(n,p) P (X ≤ 7) Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x P (X = 0)+ P (X = 1)+ P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) = 0.0000 + 0.0000 + 0.0000 +0.0004 + 0.0024 + o.0115 + 0.0401 + 0.1032 = 0.1576

  10. The probability of Sufian making at most 9 basket out of the 12 free throws is 61 % Suppose we need to find out the probability of Peter making at most 9 shots during the game, how are you going to compute for the probability of this event happening? b(n,p) P (X ≤ 9) Syntax: 2nd > VARS >arrow down (binomcdf)>n,p,x = 0.6093

  11. What if we need to find Peter’s probability of making at least 9 shots in a game? b(n,p) P (X ≥ 9) = 1- binomcdf(n,p,x) = 1- binomcdf(12,.75,9) =.3907

  12. = √n(p)( 1 - P ) μx σ2x σx Compute for the following statistical measure for this particular event: =n(p) =n(p)( 1 - p ) =9 =3.6 =2.25 =2.52 =1.5 =1.6

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